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Math Help - Help with the use of parameters with simultaneous equations

  1. #1
    Junior Member 22upon7's Avatar
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    Question Help with the use of parameters with simultaneous equations

    I have the answers to the questions, but I don't understand how to work it out, or what the answer actually means.

    The system of equations

    x+y+z+w=4
    x+3y+3z=2
    x+y+2z-w=6

    has infinitely many solutions. Describe this family of solutions and give the unique solution when w=6.


    The answers are:

    z=t

    y=\frac{-3(t+2)}{4}

    x=\frac{26-3t}{4}

    x=\frac{t-2}{2}

    (I don't know how to work out these solutions, or what it means)

    and as such, when w=6 (This bit I do know how to work out, but I don't understand what it means)

    x=-4, y=-12, z=14

    Any help would be greatly appreciated!

    Thanks!

    Drew
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by 22upon7 View Post
    I have the answers to the questions, but I don't understand how to work it out, or what the answer actually means.

    The system of equations

    x+y+z+w=4
    x+3y+3z=2
    x+y+2z-w=6

    has infinitely many solutions. Describe this family of solutions and give the unique solution when w=6.
    that's because you have 4 unknowns, x,y,z and w and just 3 equations and let's say that you can express solutions of them by the "w" but than you see that for every different values of "w" there are going to be different solutions of x,y,z

    determinant of such system is :

     D= 2 , D_x= 10-3w , D_y= -3w-6 , D_z = 4w+4

    so you have :

    \displaystyle  x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}



    so when you have w=6 that means that now w in not unknown variable, so you have :

    x+y+z+6=4
    x+3y+3z=2
    x+y+2z-6=6

    x+y+z=-2
    x+3y+3z=2
    y+x+2z=12

    just solve this system of equations for x,y and z


    do you know how to do that ?
    Last edited by yeKciM; August 28th 2010 at 11:48 PM.
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  3. #3
    Junior Member 22upon7's Avatar
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    Quote Originally Posted by yeKciM View Post
    that's because you have 4 unknowns, x,y,z and w and just 3 equations and let's say that you can express solutions of them by the "w" but than you see that for every different values of "w" there are going to be different solutions of x,y,z

    so when you have w=6 that means that now w in not unknown variable, so you have :

    x+y+z+6=4
    x+3y+3z=2
    x+y+2z-6=6

    x+y+z=-2
    x+3y+3z=2
    y+x+2z=12

    just solve this system of equations for x,y and z


    do you know how to do that ?

    Thanks for the reply!

    I know how to solve that bit, but I don't understand how you Describe this family of solutions or what it means.

    How do you get these answers:

    z=t

    y=\frac{-3(t+2)}{4}

    x=\frac{26-3t}{4}

    x=\frac{t-2}{2}
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by 22upon7 View Post
    Thanks for the reply!

    I know how to solve that bit, but I don't understand how you Describe this family of solutions or what it means.

    How do you get these answers:

    z=t

    y=\frac{-3(t+2)}{4}

    x=\frac{26-3t}{4}

    x=\frac{t-2}{2}


    i edited post

    but i think there is mistake there (probably typo) with those solutions, because okay (you don't need to) like put variable "t" to substitute "z" lol unknown as unknown so they expressed all solutions (and here is mistake) for x,y,w (not two different x) don't see how else did they lost the "w"

    and to describe family of solutions, you must know how do depending on those solutions (for let's say t-2=0 -> t= 2) how does that effect the system of equations do you know that? (solutions of the system with not homogeneous equations )

    quick reminder if you forget

    1 when  D\neq 0 and if at least one of  D_x\vee D_y\vee D_z\vee  \neq 0 than solutions are unique and given

    \displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}


    2 if  D= 0 and if at least one of  D_x\vee D_y\vee D_z\vee  \neq 0 than system of equations have no solutions


    3 if  D= 0 and if  D_x= D_y= D_z= 0 than there can be :

    a) all sub determinants of D are zeros, and coefficients with the unknowns are proportional there are 2 cases :

    - all independent members of system are proportional with coefficient with unknowns than there are infinity of solutions
    - all independent members of system are not proportional with coefficient with unknowns than there are no solutions

    b) if at least one sub determinant of D are different from zero than there are infinity of solution






    Edit: or if this is something that you still didn't learn in school , than you can do it like this :

    1  x+y+z+w = 4

    2  x+3y+3z =2

    3  x+y+2z-w=6

    than just add first and third equations so that you will get rid of "w"

    1+3 ->

    1+3)  2x+2x+3z =10
    2 )  x+3y+3z = 2

    and now you again have three unknowns and just two equations so there is again infinity of solution (and now that t=z have meaning) so you get :

     2x+2y=10-3t
     x+3y=2-3t

    multiply second with (-2) and add to the first equation

     2x+2y=10-3t
     -2x-6y=6t-4

    1+ 2) ->  -4y=6+3t  \Rightarrow y= \frac {-6-3t}{4} = \frac  {-3(2+t)}{4} and so on
    Last edited by yeKciM; August 29th 2010 at 02:26 AM.
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  5. #5
    Junior Member 22upon7's Avatar
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    Quote Originally Posted by yeKciM View Post
    i edited post

    but i think there is mistake there (probably typo) with those solutions, because okay (you don't need to) like put variable "t" to substitute "z" lol unknown as unknown so they expressed all solutions (and here is mistake) for x,y,w (not two different x) don't see how else did they lost the "w"

    and to describe family of solutions, you must know how do depending on those solutions (for let's say t-2=0 -> t= 2) how does that effect the system of equations do you know that? (solutions of the system with not homogeneous equations )

    quick reminder if you forget

    1 when  D\neq 0 and if at least one of  D_x\vee D_y\vee D_z\vee  \neq 0 than solutions are unique and given

    \displaystyle x = \frac {D_x}{D} , y= \frac {D_y}{D} , z = \frac {D_z}{D}


    2 if  D= 0 and if at least one of  D_x\vee D_y\vee D_z\vee  \neq 0 than system of equations have no solutions


    3 if  D= 0 and if  D_x= D_y= D_z= 0 than there can be :

    a) all sub determinants of D are zeros, and coefficients with the unknowns are proportional there are 2 cases :

    - all independent members of system are proportional with coefficient with unknowns than there are infinity of solutions
    - all independent members of system are not proportional with coefficient with unknowns than there are no solutions

    b) if at least one sub determinant of D are different from zero than there are infinity of solution






    Edit: or if this is something that you still didn't learn in school , than you can do it like this :

    1  x+y+z+w = 4

    2  x+3y+3z =2

    3 [tex] x+y+2z-w=6[tex]

    than just add first and third equations so that you will get rid of "w"

    1+3 ->

    1+3  2x+2x+3z =10
    2  x+3y+3z = 2

    and now you again have three unknowns and just two equations so there is again infinity of solution (and now that t=z have meaning) so you get :

     2x+2y=10-3t
     x+3y=2-3t

    multiply second with (-2) and add to the first equation

     2x+2y=10-3t
     -2x-6y=6t-4

    1+ 2 ->  -4y=6+3t Rightarrow y= \frac {-6-3t}{4} = \frac  {-3(2+t)}{4} and so on
    Thanks so much! That was very helpful!
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